Problem: Find
\[\min_{y \in \mathbb{R}} \max_{0 \le x \le 1} |x^2 - xy|.\]
The graph of
\[x^2 - xy = \left( x - \frac{y}{2} \right)^2 - \frac{y^2}{4}\]is a parabola with vertex at $\left( \frac{y}{2}, -\frac{y^2}{4} \right).$

We divide into cases, based on the value of $y.$

If $y \le 0,$ then
\[|x^2 - xy| = x^2 - xy\]for $0 \le x \le 1.$  Since $x^2 - xy$ is increasing on this interval, the maximum value occurs at $x = 1,$ which is $1 - y.$

If $0 \le y \le 1,$ then
\[|x^2 - xy| = \left\{
\begin{array}{cl}
xy - x^2 & \text{for $0 \le x \le y$}, \\
x^2 - xy & \text{for $y \le x \le 1$}.
\end{array}
\right.\]Thus, for $0 \le x \le y,$ the maximum is $\frac{y^2}{4},$ and for $y \le x \le 1,$ the maximum is $1 - y.$

If $y \ge 1,$ then
\[|x^2 - xy| = xy - x^2\]for $0 \le x \le 1.$ If $1 \le y \le 2,$ then the maximum value is $\frac{y^2}{4},$ and if $y \ge 2,$ then the maximum value is $y - 1.$

For $y \le 0,$ the maximum value is $1 - y,$ which is at least 1.  For $1 \le y \le 2,$ the maximum value is $\frac{y^2}{4},$ which is at least $\frac{1}{4}.$  For $y \ge 2,$ the maximum value is $y - 1,$ which is at least 1.

For $0 \le y \le 1,$ we want to compare $\frac{y^2}{4}$ and $1 - y.$  The inequality
\[\frac{y^2}{4} \ge 1 - y\]reduces to $y^2 + 4y - 4 \ge 0.$  The solutions to $y^2 + 4y - 4 = 0$ are $-2 \pm 2 \sqrt{2}.$  Hence if $0 \le y \le -2 + 2 \sqrt{2},$ then the maximum is $1 - y,$ and if $-2 + 2 \sqrt{2} \le y \le 1,$ then the maximum is $\frac{y^2}{4}.$  Note that $1 - y$ is decreasing for $0 \le y \le -2 + 2 \sqrt{2},$ and $\frac{y^2}{4}$ is increasing for $-2 + 2 \sqrt{2} \le y \le 1,$ so the minimum value of the maximum value occurs at $y = -2 + 2 \sqrt{2},$ which is
\[1 - (-2 + 2 \sqrt{2}) = 3 - 2 \sqrt{2}.\]Since this is less than $\frac{1}{4},$ the overall minimum value is $\boxed{3 - 2 \sqrt{2}}.$